10. 校园大扫除时,甲、乙两同学用一根长为1.8m的扁担(自身重力可忽略),在两端用力抬起一袋重120N的垃圾。若甲承受的扁担的压力为80N,问:
(1)垃圾袋悬挂处离甲用力处有多远?
(2)若将垃圾袋向乙移近0.2m,则甲承受的压力减小了多少?
答案:$解:(1)F_{乙}=G-F_{甲}=120\ \text {N}-80\ \text {N}=40\ \text {N}$
$F_{甲}l_{甲}=F_{乙}l_{乙}①$
$l_{甲}+l_{乙}=1.8\ \text {m}②$
$联立①②,解得l_{甲}=0.6\ \text {m}$
$(2)l_{甲}'=0.6\ \text {m}+0.2\ \text {m}=0.8\ \text {m},l_{乙}'=1.8\ \text {m}-0.8\ \text {m}=1\ \text {m}$
$F_{甲}'×1.8\ \text {m}=120\ \text {N}×1\ \text {m}$
$解得F_{甲}'≈66.7\ \text {N}$
$减小了∆F_{甲}=80\ \text {N}-66.7\ \text {N}=13.3\ \text {N}$